For m and n rational numbers show that mx2 + (2m + n)x + 2n = 0 has rational solutions.

Question

i can't find the discriminant worked answer would be much appreciated as i don't understand this and is hard

Answer 1

mx² + (2m+n)x + 2n = 0

 

x = [-(2m+n) ±√((2m+n)²-4(m)(2n))]/2m

 

Look at the discriminant:

(2m+n)²-8mn = 4m²+4mn+n²-8mn

= n²-4mn+4m²

This is a perfect square 

= (n-2m)²

 

The square root of this is n-2m giving us:

 

x = [-2m-n ± (n-2m)]/2m

 

x = (-2m-n+n-2m)/2m = -4m/2m = -2

 

Or

 

x = (-2m-n-n+2m)/2m = -2n/2m = -n/m

 

Obviously -2 is a rational number.

 

-n/m is a fraction with a rational number in both

numerator and denominator.  This can be expressed

as a fraction and is therefore rational.